If we denote the Cartesian coordinate of a -dimensional Euclidean space by a -vector , then a -ball centered at the origin with radius is the set of points that satisfy

Taking this definition, a 1-ball is the line segment , which its “volume” is and has no surface area. A 2-ball is a circle with radius , which has the “surface” area of and “volume” of .

Let us denote the volume of an -ball with radius by , and consider that -ball is defined by

Obviously

Therefore, we can find recursively:

## Proportionality of Volume

Firstly, we prove is proportional to by induction:

Consider , . If we have , then

Thus

## Formula for Volume

If we denote for ,

then

and

valid for if we specially define , and which

where the Gamma function is defined as

with the property that , and for positive integers . But we can also consider without resolving to the Gamma function. Consider that

It is easy to see that , , , and

thus

and

Therefore we have

Indeed, we can also prove by induction that

And the surface area of the -ball is obtained by .

## Alternative Derivation

Below is an alternative derivation by Greg Huber (1982). “Gamma function derivation of n-sphere volumes”. Am. Math. Monthly 89(5):301–302 (PDF)

Consider

for the reason that .

Denote the surface of -ball by , we can write the volume as the integration of surface area as the radius expands from zero:

or the volume as the integration of all the points in the ball:

Consider the density integration on an infinitely large -ball, where the density at point is with defined by

Using the first integral we have (assumed radius 1)

But using the second integral we have

Because they shall be equal, we have

We can also verify that .

## Properties of n-ball

If we consider the -ball with unit diameter instead of unit radius, its volume is . We can see that, due to the exponential nature in the denominator,

Therefore, the higher the dimension, the denser we can pack the balls. I first learned this result from Professor Raymond Yeung.